Find F '(x) For Each Of Thefollowing...

Anonymous

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Find f '(x) for each of thefollowing f a) f(x) = sin3 (sin2(sin x)) b) sin (sin(sin(sin(sin x))))

 

Answer No.1

DedicatedJay7

From: US
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Here is what I've got: a)3sin2(sin2(sinx))cos(sin2(sin(x)))2sin(sin(x))cos(sin(x))cos(x) b)cos(sin(sin(sin(sin(x)))))cos(sin(sin(sin(x))))cos(sin(sin(x)))cos(sin(x))cos(x)

Answer No.2

Anonymous

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Rewrite A: sin(sin(sinx)2)3 Therefore, first differentiate the outside 3(sin(sinx)2)2 Then, differentiate the next 2sin(sinx) And finally the inside cosx put them all together you'll have: 3(sin(sinx)2)22(sin(sinx))(cosx) = 6cosx(sin2(sinx))(sin(sinx))

Answer No.3

JonnyMaster14

From: US
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Votes: 27
Work from the outside in... All these steps take the form of d/dx f(g(x)) = f ' g(x) * g '(x) a) f(x) = sin3 (sin2 (sin x)) =(sin(sin(sin x))2)3 f ' = 3(sin(sin(sin x))2)2 * d/dx sin(sin(sin x))2 f ' = 3(sin(sin(sinx))2)2 * cos(sin(sin x))2 * d/dx(sin(sin x))2 f ' = 3(sin(sin(sin x))2)2 * cos(sin(sinx))2 * 2(sin(sin x)) * d/dx sin(sin x) f ' = 3(sin(sin(sin x))2)2 * cos(sin(sinx))2 * 2(sin(sin x)) * cos(sin x) * d/dx sin x f ' = 3(sin(sin(sin x))2)2 * cos(sin(sinx))2 * 2(sin(sin x)) * cos(sin x) * cos x b) f(x) = sin (sin(sin(sin(sinx)))) just like above except now there is no power rule to worryabout... f ' = cos(sin(sin(sin(sin x)))) *d/dx sin(sin(sin(sin x))) f ' = cos(sin(sin(sin(sin x)))) * cos(sin(sin(sin x))) * d/dx sin(sin(sin x)) f ' = cos(sin(sin(sin(sin x)))) * cos(sin(sin(sin x))) * cos(sin(sin x)) * d/dx sin(sin x) f '= cos(sin(sin(sin(sin x)))) * cos(sin(sin(sin x))) * cos(sin(sin x)) * cos(sin x) * d/dx sin x f ' = cos(sin(sin(sin(sin x)))) * cos(sin(sin(sin x))) * cos(sin(sin x)) * cos(sin x) * cos x Hopefully you can see the pattern here, good luck!

Answer No.4

Anonymous

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A) f(x) = sin3 (sin2 (sinx)) f '(x) = 3cos2 (sin2(sinx)) *(sin2(sinx))' =3cos2 (sin2(sinx)) * 2cos(sinx) *(sinx)' =3cos2 (sin2(sinx)) * 2cos(sinx) * cosx = 6cos2 (sin2(sinx)) * cos(sinx)*cosx b) f(x) = sin (sin(sin(sin(sin x)))) f '(x) = cos(sin(sin(sin(sinx)))) * (sin(sin(sin(sinx))))' =cos(sin(sin(sin(sinx)))) * cos(sin(sin(sinx))) * (sin(sin(sinx)))' =cos(sin(sin(sin(sinx)))) * cos(sin(sin(sinx))) * cos(sin(sinx)) *(sin(sinx))' =cos(sin(sin(sin(sinx)))) * cos(sin(sin(sinx))) * cos(sin(sinx)) *cos(sinx) * (sinx)' = cos(sin(sin(sin(sinx)))) * cos(sin(sin(sinx))) *cos(sin(sinx)) * cos(sinx) * cosx

Answer No.5

Anonymous

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This is classic chain rule. You get a present, open it and get a box. You open a box to findanother box. And another box, and another box, until there are nomore boxes to open. f(x)=sin3(sin2(sin(x))) Then f '(x)=3cos(sin2(sin(x))) * 2cos(sin(x)) * cos(x) =6 * cos(sin2(sin(x))) * cos(sin(x)) * cos(x) f(x)=sin(sin(sin(sin(sin(x)))) Then f '(x)=cos(sin(sin(sin(sin(x))))) * cos(sin(sin(sin(x)))) *cos(sin(sin(x))) * cos(sin(x)) * cos(x) And there you have it.

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